Problem: Graph this system of equations and solve. $10x-5y = 25$ $-12x+10y = -10$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $\llap{-}2$ $\llap{-}3$ $\llap{-}4$ $\llap{-}5$ $\llap{-}6$ $\llap{-}7$ $\llap{-}8$ $\llap{-}9$ $\llap{-}10$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $\llap{-}2$ $\llap{-}3$ $\llap{-}4$ $\llap{-}5$ $\llap{-}6$ $\llap{-}7$ $\llap{-}8$ $\llap{-}9$ $\llap{-}10$ Click and drag the points to move the lines.
Answer: Convert the first equation, $10x-5y = 25$ , to slope-intercept form. $y = 2 x - 5$ The y-intercept for the first equation is $-5$ , so the first line must pass through the point $(0, -5)$ The slope for the first equation is $2$ . Remember that the slope tells you rise over run. So in this case for every $2$ positions you move up $1$ position to the right. $2$ positions up from $(0, -5)$ is $(1, -3)$ Graph the blue line so it passes through $(0, -5)$ and $(1, -3)$ Convert the second equation, $-12x+10y = -10$ , to slope-intercept form. $y = \dfrac{6}{5} x - 1$ The y-intercept for the second equation is $-1$ , so the second line must pass through the point $(0, -1)$ The slope for the second equation is $\dfrac{6}{5}$ . Remember that the slope tells you rise over run. So in this case for every $6$ positions you move up $5$ positions to the right. $6$ positions up from $(0, -1)$ is $(5, 5)$ Graph the green line so it passes through $(0, -1)$ and $(5, 5)$ The solution is the point where the two lines intersect. The lines intersect at $(5, 5)$.